Question

A person writes a letter to six friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that all of them are in the wrong envelope?

Answer 1

In this question, we have to find a number of ways in which six letters can be in the envelope so that all of them are in the wrong envelope. For this, we will make use of derangement which is permutation of elements in a way that no element appears in its original position. Number of derangement for n item is given by ${{D}_{n}}=n!\sum\limits_{k=0}^{n}{\dfrac{{{\left( -1 \right)}^{k}}}{k!}}$.

Complete step by step answer:
Here we are given six letters and we have to place them in an envelope such that all letters are placed in the wrong envelope. For finding such a number of ways, let us use the formula for the number of derangements. Derangement is defined as a permutational arrangement such that no elements are placed at its original position. Here, no letter should be in the original envelope. So we need to find a number of derangements.
Number of derangement for n items is given by ${{D}_{n}}=n!\sum\limits_{k=0}^{n}{\dfrac{{{\left( -1 \right)}^{k}}}{k!}}$.
Here, n = 6 so we get, ${{D}_{6}}=6!\sum\limits_{k=0}^{6}{\dfrac{{{\left( -1 \right)}^{k}}}{k!}}$.
Let us expand the summation, we get:
${{D}_{6}}=6!\left[ \dfrac{{{\left( -1 \right)}^{0}}}{0!}+\dfrac{{{\left( -1 \right)}^{1}}}{1!}+\dfrac{{{\left( -1 \right)}^{2}}}{2!}+\dfrac{{{\left( -1 \right)}^{3}}}{3!}+\dfrac{{{\left( -1 \right)}^{4}}}{4!}+\dfrac{{{\left( -1 \right)}^{5}}}{5!}+\dfrac{{{\left( -1 \right)}^{6}}}{6!} \right]$
We know that, even power absorbs the negative sign but odd power does not, so we get,
${{D}_{6}}=6!\left[ \dfrac{1}{0!}-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5!}+\dfrac{1}{6!} \right]$
Now let us calculate all the factorials. We know that 0! = 1.
Now, 1! = 1.
2! can be written as $2\times 1$ which is equal to 2.
3! can be written as $3\times 2\times 1$ which is equal to 6.
4! can be written as $4\times 3\times 2\times 1$ which is equal to 24.
5! can be written as $5\times 4\times 3\times 2\times 1$ which is equal to 120.
6! can be written as $6\times 5\times 4\times 3\times 2\times 1$ which is equal to 720.
Putting in the values we get,
${{D}_{6}}=720\left[ 1-1+\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}-\dfrac{1}{120}+\dfrac{1}{720} \right]$
Cancelling 1 with 1 we get,
${{D}_{6}}=720\left[ \dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}-\dfrac{1}{120}+\dfrac{1}{720} \right]$
Taking LCM of 720 we get,

& \left[ 720\div 2=360,720\div 6=120,720\div 24=30,720\div 120=6 \right] \\\ & {{D}_{6}}=720\left[ \dfrac{360-120+30-6+1}{720} \right] \\\ \end{aligned}$$ Cancelling 720 by 720 we get, $${{D}_{6}}=360-120+30-6+1$$ Simplifying it we get, $${{D}_{6}}=265$$ Therefore, the number of derangements of 6 items is equal to 265. **Hence, number of ways of arranging 6 letters such that none of them is in right envelope are 265.** **Note:** Students should take care of the signs while solving this sum. Students can make mistakes by taking summation from 1 rather than 0. They should know the formula of derangement. They should note that, 0! is always equal to 1. Take care while taking LCM as it involves big calculations